Đáp án:
$19)\quad B.\, 2 + 2\ln2$
$20)\quad D.\,\dfrac72$
Giải thích các bước giải:
19) $\quad I' = \displaystyle\int\limits_1^2\dfrac{f(x)}{x^2}dx$
Đặt $\begin{cases}u = f(x)\\dv = \dfrac{1}{x^2}dx\end{cases}\longrightarrow \begin{cases}du = f'(x)dx\\v = -\dfrac1x\end{cases}$
Ta được:
$\quad I' = -\dfrac{f(x)}{x}\Bigg|_1^2 + \displaystyle\int\limits_1^2\dfrac{f'(x)}{x}dx$
$\to I' = -\dfrac{f(2)}{2} + \dfrac{f(1)}{1} + \displaystyle\int\limits_1^2\dfrac{f'(x)}{x}dx$
$\to 1 = - \dfrac42 + \dfrac11 + \displaystyle\int\limits_1^2\dfrac{f'(x)}{x}dx$
$\to \displaystyle\int\limits_1^2\dfrac{f'(x)}{x}dx = 2$
Khi đó:
$\quad I = \displaystyle\int\limits_1^2\dfrac{f'(x)+2}{x}dx$
$= \displaystyle\int\limits_1^2\dfrac{f'(x)}{x}dx + 2\displaystyle\int\limits_1^2\dfrac{1}{x}dx$
$= 2 + 2\ln|x|\Bigg|_1^2$
$= 2 + 2\ln2$
20) $\quad I' = \displaystyle\int\limits_0^{\tfrac{\pi}{2}}\sin xf(x)dx$
Đặt $\begin{cases}u = f(x)\\dv = \sin xdx\end{cases}\longrightarrow\begin{cases}du = f'(x)dx\\v = -\cos x \end{cases}$
Ta được:
$\quad I' = -\cos xf(x)\Bigg|_0^{\tfrac{\pi}{2}} + \displaystyle\int\limits_0^{\tfrac{\pi}{2}}\cos xf'(x)dx$
$\to \dfrac12 = - \cos\dfrac{\pi}{2}\cdot f\left(\dfrac{\pi}{2}\right) + \cos0\cdot f(0) + \displaystyle\int\limits_0^{\tfrac{\pi}{2}}\cos xf'(x)dx$
$\to \displaystyle\int\limits_0^{\tfrac{\pi}{2}}\cos xf'(x)dx = -\dfrac12$
Khi đó:
$\quad I = \displaystyle\int\limits_0^{\tfrac{\pi}{2}}\cos x(4+f'(x))dx$
$= \displaystyle\int\limits_0^{\tfrac{\pi}{2}}\cos xf'(x)dx + 4\displaystyle\int\limits_0^{\tfrac{\pi}{2}}\cos xdx$
$= -\dfrac12 + 4\sin x\Bigg|_0^{\tfrac{\pi}{2}}$
$= \dfrac72$
